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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

Given a triangle $ABC$ , where $O$ is its circumcenter, $M$ is a midpoint of $BC$ , and $W$ is a point of the second intersection of bisector of $C$ with the circumcircle. The line parallel to $BC$ that passes through $W$ intersects $AB$ at point $K$ , so that $BK = BO$ . Find the angle $WMB$ .

Solution

Let the line that passes through $W$ parallel to $AB$ intersect the line $BC$ at point $T$ (Fig. 6). Then, $KWTB$ is a parallelogram and:
$WT = BK = BO = WO.$
Notice that $WO \perp AB$ , since $\triangle ABO$ is isosceles, and $CW$ is a bisector of $\angle BCA$ , thus $\angle OWT = 90^\circ$ . It is also clear that $\angle OMT = 90^\circ$ , thus, $OWTM$ is inscribed with diameter $OT$ . Therefore, since $WT = WO$ , $MW$ is a bisector of $\angle OMT$ , hence $\angle WMT = 45^\circ$ .

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